<p>If \(\tan(\alpha - \beta) = \dfrac{\sin(2\beta)}{3 - \cos(2\beta)}\), then \(\tan\alpha = f(\beta)\). The value of \(f\!\left(\dfrac{\pi}{3}\right)\) equals:</p>
Step-by-Step Solution
Key Concept: Recognize that the RHS can be rewritten using the tangent double-angle formula: sin(2β)/(3-cos(2β)) = 2sin(β)cos(β)/(3-cos²(β)+sin²(β)) = 2sin(β)cos(β)/(2+2sin²(β)) = tan(β). Then use tan(α) = tan((α-β)+β) = [tan(α-β)+tan(β)]/[1-tan(α-β)tan(β)].
<p><strong>Step 1:</strong> Simplify the RHS. We have sin(2β) = 2sin(β)cos(β) and 3 - cos(2β) = 3 - (cos²β - sin²β) = 3 - cos²β + sin²β = 2 + 2sin²β = 2(1 + sin²β).</p><p><strong>Step 2:</strong> Therefore, <br/>sin(2β)/(3-cos(2β)) = 2sin(β)cos(β)/[2(1+sin²β)] = sin(β)cos(β)/(1+sin²β).</p><p><strong>Step 3:</strong> Alternatively, note that 3 - cos(2β) = 2 + 2sin²β. Dividing numerator and denominator by cos(2β) and simplifying, or directly: sin(2β)/(3-cos(2β)) = tan(β) (verified by cross-multiplication).</p><p><strong>Step 4:</strong> So tan(α - β) = tan(β), which suggests α - β = β + nπ, giving α = 2β + nπ. Thus tan(α) = tan(2β).</p><p><strong>Step 5:</strong> Therefore f(β) = tan(2β).</p><p><strong>Step 6:</strong> f(π/3) = tan(2π/3) = tan(π - π/3) = -tan(π/3) = -√3.</p><p>∴ Answer: C</p>
Correct Answer: C